Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-36, use integration by parts to evaluate each integral. $$ \int t \sqrt[3]{2 t+7} d t $$

Step-by-Step Solution

Answer

\(\int t \sqrt[3]{2t+7} \, dt = \frac{3t(2t+7)^{4/3}}{8} - \frac{(2t+7)^{7/3}}{16} + C.\)

Step 1: Choose Parts for Integration by Parts

Integration by parts requires picking two functions from the integrand: one to differentiate ( u.) and one to integrate ( dv.). For the integral \( \int t \sqrt[3]{2t+7} \, dt \), choose \( u = t \) and \( dv = \sqrt[3]{2t+7} \, dt \). This choice will simplify differentiation and integration in later steps.

Step 2: Differentiate u and Integrate dv

Differentiate \( u = t \) to find \( du \):\[du = dt.\]Integrate \( dv = \sqrt[3]{2t+7} \, dt \) to find \( v \):First, set \( x = 2t+7 \), then \( dx = 2 \, dt \). Thus, \( dt = \frac{1}{2} \, dx \). Now, the integral becomes \( \int x^{1/3} \cdot \frac{1}{2} \, dx = \frac{1}{2} \int x^{1/3} \, dx \).The solution to this integral is:\[v = \frac{1}{2} \cdot \frac{x^{4/3}}{4/3} = \frac{3x^{4/3}}{8} = \frac{3(2t+7)^{4/3}}{8}.\]

Step 3: Apply the Integration by Parts Formula

The integration by parts formula is:\[\int u \, dv = uv - \int v \, du.\]Plug \( u = t \), \( du = dt \), \( v = \frac{3(2t+7)^{4/3}}{8} \) into the formula:\[\int t \sqrt[3]{2t+7} \, dt = t \left(\frac{3(2t+7)^{4/3}}{8}\right) - \int \left(\frac{3(2t+7)^{4/3}}{8}\right) dt.\]

Step 4: Solve the Remaining Integral

Simplify the remaining integrand:\[\int \frac{3(2t+7)^{4/3}}{8} \, dt = \frac{3}{8} \int (2t+7)^{4/3} \, dt.\]Use substitution to solve this integral. Let \( x = 2t + 7 \), then \( dx = 2 \, dt \), and \( dt = \frac{1}{2} \, dx \). Substitute:\[\frac{3}{8} \int x^{4/3} \cdot \frac{1}{2} \, dx = \frac{3}{16} \int x^{4/3} \, dx.\]Solve:\[\frac{3}{16} \cdot \frac{x^{7/3}}{7/3} = \frac{3}{16} \cdot \frac{3x^{7/3}}{7} = \frac{x^{7/3}}{16}.\]Substitute back \( x = 2t + 7 \):\[\frac{(2t+7)^{7/3}}{16}.\]

Step 5: Combine All Parts of the Solution

Put together all parts:\[t \cdot \frac{3(2t+7)^{4/3}}{8} - \frac{(2t+7)^{7/3}}{16} + C.\]Simplify:\[\frac{3t(2t+7)^{4/3}}{8} - \frac{(2t+7)^{7/3}}{16} + C.\]

Step 6: Final Answer

The evaluated integral, using integration by parts, is:\[\int t \sqrt[3]{2t+7} \, dt = \frac{3t(2t+7)^{4/3}}{8} - \frac{(2t+7)^{7/3}}{16} + C.\]

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus. They calculate the net area under a curve within a particular interval. To perform a definite integral, you need to evaluate the integral expression at the upper and lower limits and then subtract the two results. This process gives you the exact area encompassed by the curve over that interval.

It is essential to understand the difference between definite and indefinite integrals. Definite integrals compute areas, while indefinite integrals result in a family of functions. This distinction is crucial in applications ranging from physics to economics where specific interval measurements are required.

Indefinite Integrals

Indefinite integrals, often referred to as antiderivatives, do not have limits of integration. They represent a family of functions rather than a specific value. Every indefinite integral comes with a constant of integration, denoted as 'C', to account for the fact that the derivative of a constant is zero.

The indefinite integral of a function is the reverse process of differentiation. Given a derivative, the indefinite integral finds the original function before differentiation. This is vital in solving differential equations and modeling dynamic systems where the original function is needed.

Substitution Method

The substitution method simplifies the integration process by transforming a complex integral into an easier one. It involves substituting part of the expression with a single variable to make the integral more manageable.

Here’s how it works:

• Identify a part of the integrand to substitute with a new variable.
• Find the differential of this new variable.
• Replace the original variable and differential in the integral.
• Integrate the new expression.
• Substitute back in the original terms after integration.

The substitution method is a powerful tool, often used in combination with other methods like integration by parts, to tackle challenging integrals.

Differentiation

Differentiation is the process of finding the derivative of a function. A derivative represents the rate at which a function is changing at any given point. It is a foundational tool in calculus, used to find slopes of curves, velocities, and a host of other rates of change within different contexts.

Differentiation has simple rules:

• The power rule allows you to differentiate polynomials easily.
• The product rule is used when differentiating two multiplied functions.
• The chain rule handles composite functions.

Understanding these rules enables you to tackle more complex calculus problems, such as those involving integration by parts and substitution methods. Knowing how to differentiate efficiently sets a good foundation for understanding integrals, as integration is essentially the reverse process of differentiation.